Q. No.A particle is executing simple harmonic motion. Its amplitude is \(A\) and time period is \(5~\text{s}\). The time required by it to move from \(x=A \text { to } x=\dfrac{A}{\sqrt{2}}\) is: (in s)
1. \(1/4\)
2. \(5/4\)
3. \(5/8\)
4. \(3/8\)
1. \(1/4\)
2. \(5/4\)
3. \(5/8\)
4. \(3/8\)
Subtopic: Angular SHM |
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Using a simple pendulum experiment \(g\) is determined by measuring its time period \(T\). Which of the following plots represent the correct relation between the pendulum length \(L\) and time period \(T\)?
| 1. |  |
2. |  |
| 3. |  |
4. |  |
Subtopic: Angular SHM |
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A simple pendulum of string length \(30\) cm performs \(20\) oscillations in \(10~\text{s}\). The length of the string required for the pendulum to perform \(40\) oscillations in the same time duration is: (in cm) [Assume that the mass of the pendulum remains same.]
1. \(120\)
2. \(0.75\)
3. \(7.5\)
4. \(15\)
1. \(120\)
2. \(0.75\)
3. \(7.5\)
4. \(15\)
Subtopic: Angular SHM |
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Two simple pendulums having lengths \(l_1\) and \(l_2\) with negligible string mass undergo angular displacements \(\theta_1\) and \(\theta_2,\), from their mean positions, respectively. If the angular accelerations of both pendulums are same, then which expression is correct?
1. \(\theta_1 l_2^2=\theta_2 l_1^2\)
2. \(\theta_1 l_2=\theta_2 l_1\)
3. \(\theta_1 l_1=\theta_2 l_2\)
4. \(\theta_1 l_1^2=\theta_2 l_2^2\)
1. \(\theta_1 l_2^2=\theta_2 l_1^2\)
2. \(\theta_1 l_2=\theta_2 l_1\)
3. \(\theta_1 l_1=\theta_2 l_2\)
4. \(\theta_1 l_1^2=\theta_2 l_2^2\)
Subtopic: Angular SHM |
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A pendulum bob is released from rest at an angle \(\theta\) with the vertical, as shown in the figure. If the magnitude of its acceleration at maximum amplitude is the same as the magnitude of its acceleration at the mean position, what is the value of \(\theta \text{?}\)
| 1. | \(\tan ^{-1}(\sqrt{2})\) | 2. | \(\text 2 \tan ^{-1}\left(\dfrac{1}{\sqrt{5}}\right)\) |
| 3. | \(2 \tan ^{-1}\left(\dfrac{1}{2}\right)\) | 4. | \(\tan ^{-1}(2)\) |
Subtopic: Angular SHM |
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Which of the following graphs best represents the relation between the square of the time period and the length of a simple pendulum?
| 1. |  |
2. |  |
| 3. |  |
4. |  |
Subtopic: Angular SHM |
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The time period of a simple pendulum in a stationary lift is \(T.\) If the lift accelerates upward with an acceleration of \(\dfrac g 6\) (where \(g\) is the acceleration due to gravity), then the time period of the pendulum would be:
| 1. | \(\sqrt{\dfrac{6}{5}} ~T \) | 2. | \(\sqrt{\dfrac{5}{6}} ~T\) |
| 3. | \(\sqrt{\dfrac{6}{7}}~T\) | 4. | \(\sqrt{\dfrac{7}{6}} ~T\) |
Subtopic: Angular SHM |
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The metallic bob of a simple pendulum has a relative density equal to \(5. \) The time period of this pendulum is \(10~\mathrm{s}. \) If the metallic bob is immersed in water, then the new time period becomes \(5 \sqrt x ~\mathrm{s}. \) The value of \(x\) will be:
1. \(5\)
2. \(7\)
3. \(9\)
4. \(2\)
1. \(5\)
2. \(7\)
3. \(9\)
4. \(2\)
Subtopic: Angular SHM |
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The time period of oscillation of a simple pendulum of length \(L,\) suspended from the roof of a vehicle, which moves without friction down an inclined plane of inclination \(\alpha,\) is given by:
| 1. | \(2\pi \sqrt{\dfrac{L}{g~ \mathrm{cos}\alpha}}\) | 2. | \(2\pi \sqrt{\dfrac{L}{g~ \mathrm{sin}\alpha}}\) |
| 3. | \(2\pi \sqrt{\dfrac{L}{g}}\) | 4. | \(2\pi \sqrt{\dfrac{L}{g~ \mathrm{tan}\alpha}}\) |
Subtopic: Angular SHM |
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The motion of a simple pendulum executing simple harmonic motion is represented by the equation;
\({y}={A} \sin (\pi {t}+\phi),\) where time is measured in seconds. The length of the pendulum is:
1. \(97.23~\text{cm}\)
2. \(25.3~\text{cm}\)
3. \(99.4~\text{cm}\)
4. \(406.1~\text{cm}\)
\({y}={A} \sin (\pi {t}+\phi),\) where time is measured in seconds. The length of the pendulum is:
1. \(97.23~\text{cm}\)
2. \(25.3~\text{cm}\)
3. \(99.4~\text{cm}\)
4. \(406.1~\text{cm}\)
Subtopic: Angular SHM |
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